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      10-05-2021, 07:18 PM   #34453
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,
Nailed it!
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      10-05-2021, 07:32 PM   #34454
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Originally Posted by Lady Jane View Post
Now I want to see the wrench.



Attachment 2712379
those guys have HUGE Nuts
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      10-05-2021, 07:54 PM   #34455
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Quote:
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Quote:
Originally Posted by Lady Jane View Post
Now I want to see the wrench.



Attachment 2712379
those guys have HUGE Nuts
.... two drums and a symbol fall off the roof
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      10-05-2021, 08:05 PM   #34456
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      10-05-2021, 08:42 PM   #34457
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Quote:
Originally Posted by shoei View Post
It actually fell from a power line but I guess velocity lol
I'm not a physicist but I'm feeling wild this evening. Let's do this....

Assumptions:
- power line height of 4.5 meters
- car windshield height of 1.4 meters
- goose weight of 6.5 kg
- the goose caused a 5cm deep indentation into the windshield

We will calculate a couple things:
- speed upon impact
- force of impact upon windshield

V = sqrt(2*g*h)
Where:
V = velocity at impact
g = acceleration (in this case gravity)
h = distance (in this case height)

So... v = sqrt(2*9.8*3.1) = 7.795 m/s = 17.4 mph

Next...

F = (m*g*h)/d
Where:
F = force
M = mass
G = gravity
H = height
D = distance (in this case the distance the goose moved the windshield as a result of the impact)

So... F = (6.5*9.8*3.1)/0.05 = 3,949 N = 887 lbs

Upon further research, the average force required to break a car windshield is 9,400 psi.

So what's the lesson here? Absolutely nothing.
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      10-05-2021, 08:50 PM   #34458
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Quote:
Originally Posted by shoei View Post
How you know to pack it in and call it a day.
That windshield is getting smashed on Grey Goose.
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      10-05-2021, 09:13 PM   #34459
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Quote:
Originally Posted by wtwo3 View Post
I'm not a physicist but I'm feeling wild this evening. Let's do this....

Assumptions:
- power line height of 4.5 meters
- car windshield height of 1.4 meters
- goose weight of 6.5 kg
- the goose caused a 5cm deep indentation into the windshield

We will calculate a couple things:
- speed upon impact
- force of impact upon windshield

V = sqrt(2*g*h)
Where:
V = velocity at impact
g = acceleration (in this case gravity)
h = distance (in this case height)

So... v = sqrt(2*9.8*3.1) = 7.795 m/s = 17.4 mph

Next...

F = (m*g*h)/d
Where:
F = force
M = mass
G = gravity
H = height
D = distance (in this case the distance the goose moved the windshield as a result of the impact)

So... F = (6.5*9.8*3.1)/0.05 = 3,949 N = 887 lbs

Upon further research, the average force required to break a car windshield is 9,400 psi.

So what's the lesson here? Absolutely nothing.
Murf993 Sounds like the rum is talking here...
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      10-05-2021, 10:29 PM   #34460
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      10-05-2021, 10:31 PM   #34461
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Quote:
Originally Posted by kscarrol View Post
Murf993 Sounds like the rum is talking here...
You beat me to it.
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      10-05-2021, 10:34 PM   #34462
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Quote:
Originally Posted by wtwo3 View Post
I'm not a physicist but I'm feeling wild this evening. Let's do this....

Assumptions:
- power line height of 4.5 meters
- car windshield height of 1.4 meters
- goose weight of 6.5 kg
- the goose caused a 5cm deep indentation into the windshield

We will calculate a couple things:
- speed upon impact
- force of impact upon windshield

V = sqrt(2*g*h)
Where:
V = velocity at impact
g = acceleration (in this case gravity)
h = distance (in this case height)

So... v = sqrt(2*9.8*3.1) = 7.795 m/s = 17.4 mph

Next...

F = (m*g*h)/d
Where:
F = force
M = mass
G = gravity
H = height
D = distance (in this case the distance the goose moved the windshield as a result of the impact)

So... F = (6.5*9.8*3.1)/0.05 = 3,949 N = 887 lbs

Upon further research, the average force required to break a car windshield is 9,400 psi.

So what's the lesson here? Absolutely nothing.
I'm going to suggest that the goose was hit by the moving vehicle for two reasons, I've never seen a goose sitting on hydro wires or in a tree for that matter and two I've seen a goose go through a car windshield before.
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      10-05-2021, 10:40 PM   #34463
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Quote:
Originally Posted by Murf993 View Post
Quote:
Originally Posted by wtwo3 View Post
I'm not a physicist but I'm feeling wild this evening. Let's do this....

Assumptions:
- power line height of 4.5 meters
- car windshield height of 1.4 meters
- goose weight of 6.5 kg
- the goose caused a 5cm deep indentation into the windshield

We will calculate a couple things:
- speed upon impact
- force of impact upon windshield

V = sqrt(2*g*h)
Where:
V = velocity at impact
g = acceleration (in this case gravity)
h = distance (in this case height)

So... v = sqrt(2*9.8*3.1) = 7.795 m/s = 17.4 mph

Next...

F = (m*g*h)/d
Where:
F = force
M = mass
G = gravity
H = height
D = distance (in this case the distance the goose moved the windshield as a result of the impact)

So... F = (6.5*9.8*3.1)/0.05 = 3,949 N = 887 lbs

Upon further research, the average force required to break a car windshield is 9,400 psi.

So what's the lesson here? Absolutely nothing.
I'm going to suggest that the goose was hit by the moving vehicle for two reasons, I've never seen a goose sitting on hydro wires or in a tree for that matter and two I've seen a goose go through a car windshield before.
The internet description said it hit a live wire and plummeted to the ground. Or rather, windshield.
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      10-05-2021, 10:46 PM   #34464
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Bird would need to grounded somehow so don’t see how a flying goose could be electrocuted…. Not to mention the associated,burns. Unless the bird broke its neck and fell. Might need another drink to consider this!
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Last edited by kscarrol; 10-05-2021 at 10:55 PM..
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      10-05-2021, 11:43 PM   #34465
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We will use the Impulse/Momentum Theorem. F
⋅∆𝑡=𝑚(𝑣𝑓−𝑣&#119894


t
=
m
(
v
f

v
i
)
Let us assume that the bird is traveling head on to the car at a speed of +12m/s=Vi.
Let us assume the bird has a mass m=0.5kg
Lets us assume the bird after impact(the impulse) is now traveling (at least for a moment) -28m/s(-60mph)=Vf. In the opposite direction it was going.
Let us assume that the time during impulse is ∆t=0.010s.
Then the average force is F=0.5kg(-28m/s-12m/s)/(0.01s)=2,000Newtons.
Most birds that would hit a windshield would be less than .5kg.
Most impacts would not result in an opposite velocity vector i.e it would bounce off the windshield at an oblique angle.
If the last two sentences applied the force would be less.
If the time of impulse is less or the initial velocity of the bird is greater then the force increases.
2,000 N distributed over an area of impact of say 100cm

2

2
would result in 20N per sq. cm. about two kilograms of weight pressing on every sq. cm of glass. I do not believe that is enough to shatter the windshield. However, I do believe that the numbers I assumed are not unreasonable for the typical bird that might hit a car. The number that is highly suspect is my impact time, it could be less than that.
The poster can play with a greater mass or a greater initial velocity or a lesser time to increase the average force. Remember, this is just the average force there is actually a greater peak force occurring at the mid-time of impact. That is harder to determine without more measurements. Assume the peak force is 4000 N.
What I could find on the pressure to break a windshield was anywhere from 4000–16000 N/cm^2 and that depends on how long the pressure is applied to the glass.
Under the conditions I assumed I do not believe the windshield would shatter.
You would need a bigger and faster bird to get a shattered windshield.
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      10-06-2021, 04:22 AM   #34466
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      10-06-2021, 05:58 AM   #34467
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Quote:
Originally Posted by shoei View Post
The internet description said it hit a live wire and plummeted to the ground. Or rather, windshield.
Disregard my comment then.
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      10-06-2021, 06:03 AM   #34468
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Quote:
Originally Posted by StripclubDJ View Post
We will use the Impulse/Momentum Theorem. F
⋅∆𝑡=𝑚(𝑣𝑓−𝑣&#119894


t
=
m
(
v
f

v
i
)
Let us assume that the bird is traveling head on to the car at a speed of +12m/s=Vi.
Let us assume the bird has a mass m=0.5kg
Lets us assume the bird after impact(the impulse) is now traveling (at least for a moment) -28m/s(-60mph)=Vf. In the opposite direction it was going.
Let us assume that the time during impulse is ∆t=0.010s.
Then the average force is F=0.5kg(-28m/s-12m/s)/(0.01s)=2,000Newtons.
Most birds that would hit a windshield would be less than .5kg.
Most impacts would not result in an opposite velocity vector i.e it would bounce off the windshield at an oblique angle.
If the last two sentences applied the force would be less.
If the time of impulse is less or the initial velocity of the bird is greater then the force increases.
2,000 N distributed over an area of impact of say 100cm

2

2
would result in 20N per sq. cm. about two kilograms of weight pressing on every sq. cm of glass. I do not believe that is enough to shatter the windshield. However, I do believe that the numbers I assumed are not unreasonable for the typical bird that might hit a car. The number that is highly suspect is my impact time, it could be less than that.
The poster can play with a greater mass or a greater initial velocity or a lesser time to increase the average force. Remember, this is just the average force there is actually a greater peak force occurring at the mid-time of impact. That is harder to determine without more measurements. Assume the peak force is 4000 N.
What I could find on the pressure to break a windshield was anywhere from 4000–16000 N/cm^2 and that depends on how long the pressure is applied to the glass.
Under the conditions I assumed I do not believe the windshield would shatter.
You would need a bigger and faster bird to get a shattered windshield.
That goose is likely a lot more than .5KG.

From Wiki:

The male Canada goose usually weighs 2.6–6.5 kg (5.7–14.3 lb), averaging amongst all subspecies 3.9 kg (8.6 lb). The female looks virtually identical, but is slightly lighter at 2.4–5.5 kg (5.3–12.1 lb), averaging amongst all subspecies 3.6 kg (7.9 lb), and generally 10% smaller in linear dimensions than the male counterparts
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      10-06-2021, 06:47 AM   #34469
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Last edited by Lady Jane; 01-06-2022 at 06:29 PM..
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      10-06-2021, 07:25 AM   #34470
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Quote:
Originally Posted by shoei View Post
How you know to pack it in and call it a day.
Can't help but think about that WKRP episode.
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      10-06-2021, 07:48 AM   #34471
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Quote:
Originally Posted by shoei View Post
.
The short video clip of this is hilarious. Ole boy on the left eyeballing Snoop as he puffs away, then Snoop offers it to the officer, the officer semi-declines, Snoop persists aggressively, then the officer goes to town on it. I mean, how do you tell Snoop no and how to do you pass up the opportunity to smoke with him?
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      10-06-2021, 08:09 AM   #34472
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Quote:
Originally Posted by King Rudi View Post
The short video clip of this is hilarious. Ole boy on the left eyeballing Snoop as he puffs away, then Snoop offers it to the officer, the officer semi-declines, Snoop persists aggressively, then the officer goes to town on it. I mean, how do you tell Snoop no and how to do you pass up the opportunity to smoke with him?
A buddy of mine shared a joint with Ziggy Marley....I mean how do you say no to that.
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      10-06-2021, 08:28 AM   #34473
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      10-06-2021, 08:58 AM   #34474
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Quote:
Originally Posted by wtwo3 View Post
I'm not a physicist but I'm feeling wild this evening. Let's do this....

Assumptions:
- power line height of 4.5 meters
- car windshield height of 1.4 meters
- goose weight of 6.5 kg
- the goose caused a 5cm deep indentation into the windshield

We will calculate a couple things:
- speed upon impact
- force of impact upon windshield

V = sqrt(2*g*h)
Where:
V = velocity at impact
g = acceleration (in this case gravity)
h = distance (in this case height)

So... v = sqrt(2*9.8*3.1) = 7.795 m/s = 17.4 mph

Next...

F = (m*g*h)/d
Where:
F = force
M = mass
G = gravity
H = height
D = distance (in this case the distance the goose moved the windshield as a result of the impact)

So... F = (6.5*9.8*3.1)/0.05 = 3,949 N = 887 lbs

Upon further research, the average force required to break a car windshield is 9,400 psi.

So what's the lesson here? Absolutely nothing.
Well done; however, there simply isn't enough information to deduct any of this. I do like where your head is at, as I instantly started calculating in my head as well. Here are my thoughts:

Power line height you have listed as 4.5 meters, yet the NECs industry standard minimum for attachment points on a telephone pole is 18 feet on line lengths up to 20 feet. Roughly 5' variance, not much, but considering the weight of the Canada Goose, this will greatly impact velocity. Also another point of consideration is random heights of power lines. Although industry standards are set, they are rarely often adhered to. Also, pole height and attachment clearance regulation vary depending on the location of the pole itself. Thus creating an undeterminable height from whence the creature fell, also thus making velocity inconclusive, which pretty much destroys all calculations after.

Also, car windshield height that you have listed seems accurate; the grade (downward slope/angle) of said windshield needs to be accounted for as well. The grade of the windshield will also effect displacement of force upon impact.

You have calculated the goose created a 5cm (= >2") impression in the windshield, I would imagine that this figure is conservative at best based off the photo.

Next in order to accurately calculate velocity, the distance is extremely important and without knowing the exact distance or point of origin of acceleration, it is difficult to determine whether or not this creature reached terminal velocity. Also further considerations must include drag coefficient, projected area and mass of the falling object; but it looks like you covered this in force.

I was about to go down the gambit of info in order to attempt to calculate what I could from this, but there simply isn't enough data. I will reverse engineer the shit out of this so stand back.

if a windshield has a tensile strength of 9,400 psi this would equate to 64810718.5 N/m˛. This raises a plethora of questions for me now that can not be answered based solely off the picture. 1.) At what speed was the vehicle traversing during impact, 2.) What was the internal temperature and external temperature of the windshield, 3.) At what trajectory or angle did this impact take place and finally, what hue of brown/consistency of mass was the excrement removed from the operators underpants post impact.
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